Why is $\mathbb{C}$ called algebraically closed? [closed]

I understand that polynomials like $x^2+1=0$ can only be solved using complex numbers. But what about other operations like $\sqrt{x}+1=0$ (or if this equation has a solution, why isnt't Wolfram Alpha able to solve it).

• Possible duplicate of Why is it that Complex Numbers are algebraically closed? – emka Jan 10 '17 at 20:51
• Is your question about $\mathbb{C}$ in general, or about why Wolfram Alpha can't solve that equation? I'm not sure if Mathematica stackexchange covers questions about Wolfram Alpha (mathematica.stackexchange.com) but if this question doesn't get an answer here, then Mathematica stack exchange might also be worth trying – Chill2Macht Jan 10 '17 at 20:53
• For $\sqrt{x}=-1$ see this duplicate. – Dietrich Burde Jan 10 '17 at 20:55
• The convention Wolfram Alpha uses for $\sqrt{z}$ is to choose the solution to $x^2 = z$ with positive real part. With this convention, $\sqrt{z} + 1 = 0$ is impossible. – eyeballfrog Jan 10 '17 at 21:05

For the term "algebraically closed" we specifically mean that it contains all zeroes of polynomial equations. In other words, equations which may be written only by using complex numbers, the unknown $x$, multiplication and addition. $\sqrt{{}\cdot{}}$ is not one of the allowed operations in this context. Neither is exponentiation (except ones that may be written out as multiplication, i.e. positive integer exponents), logarithms, or division (although you can multiply with any complex number you want, for instance $\frac15$).
If your equation can be written using these rules, then the equation has a solution among the complex numbers. That's the fundamental theorem of algebra, also known as "$\Bbb C$ is algebraically closed"